Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 314 / 1

A magnetostatic force-free magnetic field is in equilibrium when its electric current exerts no Lorentz force density:Thus $\mathbf{J}$ is pointwise parallel to $\mathbf{B}$, so for some scalar $\alpha (\mathbf{r})$,Using the magnetostatic Ampère's law, $\nabla \times \mathbf{B}=(4\pi /c)\mathbf{J}$, givesTaking the divergence and using $\nabla \cdot \mathbf{B}=0$ yieldsThe force-free parameter is therefore constant along each magnetic field line.

Put $q(x)=4\pi \alpha (x)/c$. Because every component depends only on $x$, the solenoidal constraint gives $B_x^{\prime }=0$, while the force-free equation givesFor a general nonzero, nonconstant $q$, $B_x=0$. Eliminating $B_y=-B_z^{\prime }/q$ gives the closed equationor equivalently $(B_z^{\prime }/q{)}^{\prime }+q{B}_z=0$. Since $\nabla \alpha ={\alpha }^{\prime }(x){\mathbf{e}}_x$ and $B_x=0$, the constraint $\mathbf{B}\cdot \nabla \alpha =0$ is automatically satisfied.

Define the accumulated rotation angleThe coupled first-order equations describe a rotation of $(B_y,B_z)$ and have the general solutionDirect differentiation verifies both force-free equations, and $B_y^2+B_z^2=C_1^2+C_2^2$ is constant. Thus a one-dimensional force-free magnetic field rotates without changing its magnitude.

Hereafter choosing the integration constant so that $\theta (\pm \infty )=0$. The condition $B_z\to 0$ at both ends removes the cosine solution. Up to an overall sign and magnitude, one may writeThe angle ranges from $0$ to $I_{\max }=a^2/(2d^2)$. Neither nonzero component changes sign when the entire range lies in the first quadrant, namely $I_{\max }\le \pi /2$. Thereforefor positive length scales $a,d$. Equality allows $B_y$ to touch zero at $x=0$ without changing sign.